SEPARATION SEP ARATION PROCESS PROCE SS II (CDB 2063)
Dr Nurul Aini Amran
[email protected]
RECAP
Hy (kJ/kg dry air) =( 1.005 1.005 + 1.88 H) (T - T0 ºC ) +H 0 air. K) = 1.005 cS (kJ/kg dry air.
+ 1.88 H
vH (m3/kg dry air) = (2.83 x 10 – 3 + 4.56 x 10-3 H) T K
LET’S SEE IF YOU UNDERSTAND THE CONCEPT
ASSESSMENT 1 Air entering a dryer has a temperature (dry bulb temperature) of 60ºC and a dew point of 26.7ºC. Using the humidity chart, determ determine ine the: the: a) actu actual al humi humidi dity ty,, H b) pe perc rcen enta tage ge humi humidi dity ty,, H P c) humid heat, cs d) humi humid d vo volu lume me,, vH
ASSESSMENT 1: SOL SOLUTION UTION
Percentage humidity, H P = 14% H = 0.0225 …
ASSESSMENT 1: SOL SOLUTION UTION a) Actual humidity, H Refer Humidity Chart, given Dry bulb temperature (air temperature = 60ºC, Dew Point = 26.7 ºC)
H = 0.0225 kg H2O/kg dry air b) Percentage humidity, H P Hp = 14% c) Humid heat? dry air) = 1.005 + 1.88 H Cs = 1.047 kJ/kg dry air . K cS (kJ/kg dry
d) Humid volume? vH = (2.83 x 10 – 3 + 4.56 x 10-3 H ) T K T K
VH = 0.977m3/kg dry air
ASSESSMENT 2 An air stream at 87.8ºC having a humidity H = 0.03 kg H2O/kg dry air is contacted in an adiabatic saturator with water. It is cooled and humidified to 90% saturation. Determine: a)
The final values of H and T
b)
For 100% saturation, what would be the values of H and T a. H = 0.05 kg H 2O/kg dry air, T = 42.5 ºC
What is the wet bulb temperature of this air stream?
b.
H =
0.0505 kg H2O/kg dry air,
ASSESSMENT 2: SOLUTION Given: 87.8ºC, H = 0.03 kg H2O/kg dry air, 90% saturation
a)The final values of H and T H = 0.05 kg H 2O/kg dry air, T = 42.5 ºC b)For 100% saturation, what would be the values of H and T
90%
H = 0.0505 kg H 2O/kg dry air,
H =0.05
0.03 or 3%
T = 40.5ºC
T =42.5
87.8
EXAMPLE 9.3-4 A water vapor-air mixture having a dry bulb temperature of T = 600C is passed over a wet bulb. The wet bulb temperature obtained is T w = 29.5 0C. What is the humidity of the mixture? Solution: The adiabatic saturation lines can also be used for wet bulb lines with reasonable accuracy. So T w can be assumed to be the same as T S
HUMIDITY CHART
H =0.0135 60 29.5
LESSON OUTCOME At the end of this topic, students should be able to:
describe the drying principles and its application
apply the following concept in solving problems related to drying
humidity, humid volume, humid heat
adiabatic saturation temperature
wet bulb temperature
equilibrium moisture content
constant rate and falling rate drying
differentiate the different types of drying region
predict the drying rate and drying time using empirical correlation
In pair, suggest the theory used in justifying why each of these method is better than indoor drying, and when does drying stops?
Warm air is less dens e than cold air, there is more vapor i n warm air as compared with cold air.
‘room’ for
water
WHAT WILL HAPPEN WHEN A WET SOLID IS BROUGHT INTO CONTACT WITH AIR?
EQUILIBRIUM MOISTURE CONTENT (EMC) OF MATERIALS attained when solid attain a definite moisture content after prolonged exposure to air under specified humidity and temperature of air.
Equilibrium
It’s
the point where solid stops losing moisture after being contacted with air
Changes with relative humidity and temperature
UNIT
= kg water/kg bone dry solid (dry basis) = kg moisture/kg wet substance (wet basis) = kg H2O/100 kg dry solid = kg water/kg dry solid
EQUILIBRIUM MOISTURE CONTENT OF MATERIALS
EMC varies greatly with the type of material
Nonporous insoluble solid tend to have low EMC at given humidity
Typical
food or biological material tend have high EMC
EMC of a solid decreases with increase in temperature at given humidity.
FREE MOISTURE OF A SUBSTANCE
Free moisture content is the difference between given moisture content of solid, X T and equilibrium moisture content, X*, => X T – X* ( moisture in excess to EMC)
This
excess moisture is removed by drying under the given percent relative humidity Subject to air flow (RH 50%, 25 °C
Silk, EMC = 8.5 kg H2O/100 kg dry material (RH air: 50% at 25oC)
Silk sample, water content = 10 kg H2O/100 kg dry material
THE DRYING PROCESS
Things to consider: 1. Operating condition – air temperature and air flow ( air humidity) 2. Drying time 3. EMC of material
Rate of drying curve from experiment
X t X
Sample (solid) is placed on tray
measure weight loss of moisture during drying at different intervals without interrupting the operation Constant drying condition : Velocity, humidity, temperature and direction of
W
LS
LS
X t
X *
DATA W = weight of wet solid, kg wet solid LS = kg dry solid X = free moisture content, kg free H2O/kg dry solid X* = EMC, kg equilibrium moisture/kg dry solid Xt = total water, kg total H2O/kg dry solid
DRYING CURVES Graph 1: Drying rate curve in terms of free moisture content
Graph 2: Drying rate curve in terms of drying rate
Measure the slope of tangents to the curve
Aʹ = when solid is hot A = when solid is colder
X t X
W
LS
LS X
X *
R
LS X A
t
LS dX A
dt
THE DRYING RATE
Constant rate region
• Surface of solid initially very wet (film of moisture) • Rate of evaporation is independent of solid, and similar to rate from free liquid surface • The drying rate is mostly dependent on the rate of heat transfer to the solid
Falling rate region
• Starts at point C • Insufficient water on the solid surface • Water migration from the solid interior to the surface by molecular diffusion or capillary movement. • At point D, the surface is completely dry • Plane of evaporation recedes from surface • Point C to E is a long
DRYING CURVE
CONSTANT RATE PERIOD R
R
LS X A
t
LS dX A
LS dX A
t
dt
t
L s ARC
t 2 t
L s
t 1 0
A
X 1
dt
dt
X 2
X 1 dX
X 2
R
FALLING RATE PERIOD R
Critical moisture content
E S AR C L A I C E R P S
LS dX
= aX+b
= aX
A
dt t
t
A R1
Ls X 1
L s X C ARC
X 2
R2
ln
ln
X C X 2
R1 R2
WHAT IF THE DRYING IN THE FALLING RATE PERIOD DOES NOT FALL INTO ANY OF THE SPECIAL CASE? t
LS A
X 1
X 2
dX R
WHAT IF EXPERIMENTAL DATA IS NOT AVAILABLE? Use equation to predict the constant-rate drying
Assumptions: i. Steady state process i.e. rate of mass transfer balances the rate of heat transfer ii. Neglect heat transfer by conduction and radiation iii. Heat transfer is by air convection only Convective heat transfer from gas to surface of solid:
q
h(T T W ) A
Heat needed to vaporize NA (kmol/s.m 2) water:
q M A N A W A
Mass flux of water vapor from the surface:
N
k ( y
y) k
M B
( H
H )
WHAT IF EXPERIMENTAL DATA IS NOT AVAILABLE? RC
q
h(T T W )
A W
W
k y M air ( H W H )
Drying time for constant rate drying (in terms of transfer coefficients)
t
LS W ( X 1 X 2 )
Ah(T T W )
where h is predicted using:
in °C
LS ( X 1 X 2 )
Ak y M air ( H W H )
2
Rc (kgH 2O / h.m )
h = 0.0204 G0.8
h(T T W )
W
3600
G, mass velocity = 2450 – 29 300 kg/h m2 Velocity of air = 0.61 – 7.6 m/s
h = 1.17 G 0.37
G, mass velocity = 3900 – 19 500 kg/h m2
GROUP ACTIVITY
CALCULATION FOR DRYING TIME IN THE CONSTANT RATE PERIOD A solid with drying curve shown below is to be dried from X1 = 0.38 kg H2O / kg dry solid to X2 = 0.25 kg H2O / kg dry solid. Estimate the time required for drying Given: free moisture content X 1 and X 2 Find: drying time Solution: From drying curve (Fig 9.5-1a), X 1 t1 and X 2 t2 Hence, drying time = t2 – t1
CALCULATION FOR DRYING TIME IN THE CONSTANT RATE PERIOD A solid with drying curve shown below is to be dried from X1 = 0.38 kg H2O / kg dry solid to X2 = 0.25 kg H2O / kg dry solid. Estimate the time required for drying
If the drying curve is not available, but given that the rate of drying in the constant rate period is 1.51 kg H2O / h.m2, and that Ls/A = 21.5 kg dry solid /m2 , calculate the drying time to dry the sample from X1 = 0.38 kg H2O / kg dry solid to X2 = 0.25 kg H2O / kg dry solid. R
LS dX A dt t t
L X dX dt s t 0 A X R Drying time for constant rate period t
2
1
t
L s ARC
1
2
X 1 X 2 1.85 h
CALCULATION FOR DRYING TIME IN THE FALLING RATE PERIOD Find the time for drying if a solid is to be dried from XC = 0.195 kg H2O / kg dry solid to X = 0.04. Take R C=1.51 kg H2O / h.m2, and Ls/A = 21.5 kg dry solid /m 2. Assume straight line through origin t
L s X C ARC 4.39 h
ln
X C X 2
CALCULATION FOR DRYING TIME IN THE FALLING RATE PERIOD Find the time for drying if a solid is to be dried from XC = 0.195 kg H2O / kg dry solid to X = 0.04. Take Ls/A = 21.5 kg dry solid /m2. Use numerical integration to solve the problem
t
X
R
LS A
1/R
X 1
X 2
dX R X
(1/R)av
X/(1/R )av
CALCULATION FOR DRYING TIME IN THE FALLING RATE PERIOD Find the time for drying if a solid is to be dried from XC = 0.195 kg H2O / kg dry solid to X = 0.04. Take Ls/A = 21.5 kg dry solid /m2. Use numerical integration to solve the problem
t
X
R
LS A
1/R
X 1
X 2
X
dX R (1/R)av
(∆X)(1/R)av
0.195
1.51
0.663
0.045 0.745
0.0335
0.150
1.21
0.826
0.050 0.969
0.0485
0.100
0.90
1.111
0.035 1.260
0.0441
0.065
0.71
1.408
0.015 2.055
0.0308
0.050
0.37
2.702
0.010 3.203
0.0320
0.040
0.27
3.704
t
LS
A
X 1
X 2
dX R
TOTAL
0.1889
21.5 0.1889 4.06h
PREPARE A MIND MAP COVERING THE LECTURE MATERIAL TODAY FOR YOUR GROUP
LECTURE 3
– TUTORIAL
SESSION
RECAP – PREVIOUS LECTURE X t
FMC
X
W
R
LS
X t
X *
CONSTANT RATE PERIOD WITHOUT EXPERIMENT DATA
CONSTANT RATE PERIOD
LS
EMC t
LS dX
L s ARC
A
X 1
dt
X 2
RC
q
A W
h(T T W )
t
FALLING RATE PERIOD
W
k y M air ( H W H )
LS W ( X 1 X 2 )
Ah(T T W )
R E S R A C L A I C R E P S
LS dX A
= aX+b
dt
t
A R1
Ls X 1
LS ( X 1 X 2 )
Ak y M air ( H W H )
X 2
R2
ln
R1 R2
h = 0.0204 G 0.8 = aX
t
L s X C ARC
ln
X C X 2
If no, use numerical integration / measure area under the graph of
h = 1.17 G
0.37
RECAP – GROUP BRAINSTORMING Discuss
how the following affects the drying process (assume no radiation and conduction heat transfer)
Share
AIR VELOCITY
AIR TEMPERATURE
AIR HUMIDITY
SOLID THICKNESS
your findings with the whole class
EXAMPLE: 9.6-3: PREDICTION CONSTANT- RATE DRYING
An insoluble wet granular material is dried in a pan 0.457 x 0.457 m, and 25.4 mm deep. The material is 25.4 mm deep in the pan, and the sides and bottom can be considered to be insulated. Heat transfer by convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s. The air is at 65.6 oC and has a humidity of 0.010 kg H2O/kg dry air. Estimate the rate of drying for the constant-rate period.
EXAMPLE :9.6-3:GIVEN Given: A = 0.457 x 0.457 m Deep = 25.4 mm deep Insulated = adiabatic Air parallel, velocity = 6.1 m/s T = 65.6 oC H = 0.010 kg H 2O/kg dry air
Find: rate of drying for the constant-rate period,
EXAMPLE :9.6-3:SOLUTION Assumption: adiabatic Rate of drying for constant-rate period
RC
h
W
(T
0
T W C )(3600) [to express the rate per hour]
h = heat transfer coefficient (W/m 2 K) T = temperature of air T W = wet bulb temperature = solid surface temperature W = latent heat at TW (J/kg)
1. T w = ? Refer Humidity Chart, T w = 28 0C, H w =0.026
H w H
T w
EXAMPLE :9.6-3:SOLUTION 2. Refer steam table for W, at T w = 280C, W = 2433 kJ/kg
3. How to obtain h, heat transfer coefficient??? h = 0.0204 G 0.8 [for parallel air flow]
EXAMPLE :9.6-3:SOLUTION Mass Velocity, G = v
Velocity of air = GIVEN
Density of air, = m/ H (Total mass of dry air + moisture)/Humid volume
vH = (2.83 x 10 – 3 + 4.56 x 10-3 H ) T K = 0.974 m3/kg dry air = m/H = 1 (kg dry air) + 0.01 (kg H2O) = 1.037 kg/m 3 0.974
G = 22 770 kg/h m 2
EXAMPLE :9.6-3:SOLUTION Then,
h = 0.0204 G0.8 = 62.45 W/m2 K
4. Substitute all values into Equation RC
h
W
(T
0
T W C )(3600)
Rc = 3.39 kg/h m2
(drying rate for constant period)
What is the total evaporation rate for a surface area of 0.457 x 0.457 m2 ?
RC A = 0.708 kg/hr
TUTORIAL QUESTION 1 A batch of wet solid was dried on a tray dryer under constant drying conditions from initial free moisture content of X1 = 0.55 to critical moisture content, X C = 0.22 kg free moisture / kg dry solid. Thickness of the material on the tray is 25.4 mm. Only the top surface is exposed to air flow. The drying rate during the constant rate period is 2.05 kg H 2O/h.m2, and Ls/A = 24.4 kg dry solid/m 2. Calculate the time to dry the same material from X 1 = 0.45 to X2 = 0.30 if the thickness of material is 50.8 mm, with both top and bottom surface exposed to air flow. (Ans: 1.785 h)
ANSWER TUTORIAL QUESTION 1 The thickness of the solid is doubled. Since both top and bottom surfaces are exposed to air, LS/A remains the same.
t
L s ARC
X 1
X 2
24.4kgdrysolid / m
3
2
2.05kgH 2O / m .h 1.785hours
(0.45 0.3)
TUTORIAL QUESTION 2 A tile company produces 1-in. thick tiles with the dimension of 1 ft 1 ft as one of their main products. The tiles are dried on the top surface using air at a wet bulb temperature of 80ºF and a dry bulb temperature of 120ºF flowing parallel to the surface at a velocity of 3.5 ft/s. The dry density of the tile is 120lb m/ft3 and the critical free moisture content is 0.09 lbm H2O/lb m dry solid. The latent heat of water at the wet bulb temperature is 1. Calculate the drying rate for the constant-rate drying region, Rc. 1048.3 btu/lbm. The humidity chart for mixtures of air and water vapor at a total pressure of 101.325 2. Calculate how long it will take to dry the tiles from a free moisture content of 0.2 lbm H2O/lbm dry kPa (760 mm Hg) is provided in FIGURE Q2
Tw = 80 °F T = 120 °F Xc = 0.09 lbm H2O/lbm dry solid W = 1048.3 btu/lbm A = 1 ft2 v = 3.5 ft/s dry = 120 lbm/ft3
ANSWER TUTORIAL QUESTION 2 At Tw = 80 °F, T = 120 °F, H = 0.013 kg lbm H 2O/lbm dry air G
= v
=
G h
m/H = ___1 (lbm dry air) + 0.013 (lbm H2O) __ = 0.0679 lbm/ft3 (0.0252 + 0.0405(0.013))(120+459.67)
= 3.5 ft/s x 0.0679 lbm/ft3 x 3600 s/h = 855.54 lbm/ft2.h
= 0.0128(855.54)0.8 = 2.838 btu/h.ft2.F
Hence,
drying rate,
h
W
(T
0
T W C )
0.108 lbm/h.ft 2
L s ARC
X
1
X 2
Ls = dry x volume tile = 120 x 1 x 1 x (1/12) = 10 lbm. Ls/A = 10 lbm/ft2
Hence, t
RC
t
10
0 108
0.2
0.09
10.19 hrs
TUTORIAL QUESTION 3 In a small textile processing unit, dyed and washed cotton yarn is dried in an insulated chamber through which hot air is blown. Bundles of yarn are kept suspended from the roof of the chamber through which hot air is blown. 200 Kg of wet yarn with 1 kg water per kg yarn is taken in a batch. Air at 25 oC (wet bulb temperature =15oC) is heated to 65 oC before it enters the chamber. Assume airflow rate is large enough for its temperature and humidity to remain constant. From the operating data it is known that the constant rate of drying under the given conditions is 0.5 kg moisture per hour per kg of dry yarn. The critical moisture is X c= 0.15, and the equilibrium moisture is X*=0.001. Calculate the time required for drying the yarn to a final moisture of Xf =0.003 if the falling rate of drying is linear in X.
ANSWER TUTORIAL QUESTION 3 X1
= Xt – X* = 1 – 0.001 = 0.999 kg H2O/kg yarn
Constant t
Xt = 1 kg H2O/kg dry yarn X* = 0.001 kg H2O/kg dry yarn Xc = 0.15 kg H2O/kg dry yarn T = 65 °C Tw = 15 °C Rc = 0.5 kg H2O/h. kg dry yarn
L s ARC
drying period,
X
1
X 2
= 0.5 kg H2O/h. kg dry yarn (0.999 – 0.15) = 1.7 hours. Falling t
drying period, assume R = ax (special case)
L s X C ARC
ln
Hence total drying period,
X C X 2
= (0.15/0.5) ln (0.15/0.003)
t = 1.7 + 1.174 = 2.874 hours.
TUTORIAL QUESTION 4 In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray drier with air flowing over the top surface with an area of 0.186 m2. The bone dry sample weight was 3.765 kg dry solid. Experimental data is given below. Evaluate the data and find the value of drying rate in the constant rate period. Next, predict the total time to dry the sample from X = 0.2 to X = 0.04. Time (h)
Weight (kg)
Time (h)
Weight (kg)
Time (h)
Weight (kg)
0
4.944
2.2
4.554
7.0
4.019
0.4
4.885
3.0
4.404
9.0
3.978
0.8
4.808
4.2
4.241
12.0
3.955
1.4
4.699
5.0
4.150
ANSWER TUTORIAL QUESTION 4 X versus t
t 0.0 0.4 0.8 1.4 2.2 3.0 4.2 5.0 7.0 9.0 12.0
W 4.944 4.885 4.808 4.699 4.554 4.404 4.241 4.15 4.019 3.978 3.955
X 0.263 0.247 0.227 0.198 0.159 0.119 0.076 0.052 0.017 0.006 0.000
dX/dt -0.039 -0.051 -0.048 -0.048 -0.050 -0.036 -0.030 -0.017 -0.005 -0.002
0.300
R
0.250
The gradient = -0.049
0.200
0.100
0.2
3.765
dt
( 0.049)
0.991 kgH 2 O/m 2 .h
0.000 0.0
2.0
4.0
t
6.0
8.0
L s ARC
X 1
X c
3.765
10.0
12.0
Constant drying period,
0.6 0.4
A
0.050
1 0.8
LS dX
0.186
0.150
R versus X 1.2
0.186(0.991) 1.62 hours
14.0
Falling t
(0.2 0.12)
drying period, assume R=ax,
L s X C ARC
ln
X C X 2
3.765 0.12
0.12 ln 0.186 0.991 0.04
2.69hours
otal drying period = 4.31 hours
TUTORIAL QUESTION 5
Assumptions: • For every batch, tea leaves are distributed evenly on a single drying tray in the drying cabinet. • Hot air from the blower flow perpendicular to the tea leaves. • Constant drying rate occurs for the production of good quality tea and the equilibrium moisture content, X * = 0.05 kg H2O/kg bone dry solid. • The drying cabinet is considered to be insulated. = 2442 • Latent heat of vaporization, λw kJ/kg. • Humid volume, vH (m3/kg dry air) = (2.83 x 10-3 + 4.56 x 10-3 H) T • Heat transfer coefficient, h (W/m2.K) = 1.17 G 0.37
For the drying process in a local black tea manufacturing company, the optimum drying period is 1.5 hours and the free moisture to be removed is 30% of the initial feed moisture content. Batch drying process is conducted inside a drying cabinet (width = 5 m, length = 5 m and height 0.5 m). Electric blowers are mounted on the upper part of the cabinet to dry the tea leaves in the drying cabinet. The initial free moisture content for tea leaves is 0.4 kg H2O/kg dry solid. The drying cabinet is operated at 51ºC and has a humidity of 0.01 kg H2O/ kg dry air. The company produces 500 kg of dry black tea per batch. 1. Calculate the required velocity of air flow for the process 2. Using the same operating conditions, estimate the amount of green tea produced (kg) per batch, if the final free moisture content of the green tea is 0.25 kg H O/kg dry solid. The initial free moisture content for
ANSWER TUTORIAL QUESTION 5 A)
Hence,
Calculate X2 X2 = 0.7Xt1 – X* = (0.7 (0.4 + 0.05)) – 0.05 = 0.265
Assume
operation in constant rate region (based on data limitation), t
Rc
L s ARC
LS
X 1
X 2
25m (1.5hr ) 1.8 kg/m 2 .h
h
(0.4 0.265)
W
(T
o
T W C )
1.8 kg/m 2 .h
At T = 51°C, H = 0.01 kg H2O/kg dry solid, Tw = 25°C, λW = 2442.31 kJ/kg
Rc W T
T W
(1.8 kg/m 2 .h)(2442.31 kJ/kg)
h=
1.17G0.37, hence G = 21584.79 kg/m2.h
G=ρv V
51 - 25
46.97 J/m 2 .o C
2
RC
h
( X 1 X 2 )
At 500kg
Ls = 500 kg dry solid t = 1.5 hours X1 = 0.4 kg H2O/kg dry solid T = 51°C H = 0.01 kg H2O/kg dry solid
= 5.53 m/s
QUESTION?
DRYING IN THE CONSTANT-RATE PERIOD
surface of solid is initially very wet
continuous film of water (unbound water) exists on drying surface
rate of evaporation & drying is independent of solid
rate same as the rate from a free liquid surface
For porous solid
most of the water is supplied from interior of solid
The
constant period continues until water supply is as fast as evaporation
Surface temperature is same as the wet bulb
C
B A
D E X
